OFFSET

1,2

COMMENTS

All terms are odd and squares (A016754).

FORMULA

a(n) = A038547(n)^2. - Thomas Scheuerle, Sep 30 2022

Proof: Suppose a(n) = Product p_i^(2*e_i), where the p_i are odd primes. Then the n odd square divisors are all of the form d = Product p_i^(2*k_i) with 0 <= k_i <= e_i. As a(n) = Product (p_i^e_i)^2 = (Product (p_i^e_i))^2, we get that sqrt(a(n)) = Product (p_i^e_i). This is the prime decomposition of sqrt(a(n)). As there is a bijection between prime factors p_i^(2*k_i) and (p_i^k_i), there is also bijection between odd square divisors of a(n) and odd divisors of sqrt(a(n)). We conclude that sqrt(a(n)) is the smallest integer that has exactly n odd divisors. - Bernard Schott, Oct 01 2022

a(p) = 3^(2*(p-1)) for primes p. - Bernard Schott, Oct 03 2022

EXAMPLE

2025 has 6 divisors that are odd squares: {1, 9, 25, 81, 225, 2025}; also, 2025 is the smallest integer that has 6 odd squares divisors, hence a(6) = 2025.

PROG

(PARI) f(n) = factorback(apply(e->e\2+1, factor(n/2^valuation(n, 2))[, 2])); \\ A298735

a(n) = my(k=1); while (f(k)!=n, k++); k; \\ Michel Marcus, Sep 29 2022

CROSSREFS

KEYWORD

nonn

AUTHOR

Bernard Schott, Sep 29 2022

EXTENSIONS

a(7)-a(10) from Michel Marcus, Sep 29 2022

More terms from Amiram Eldar, Sep 29 2022

STATUS

approved